JEE Advanced 2025 Paper 1 · Q12 · Wave Optics
A single-slit diffraction experiment uses $b\cdot d/D = m\cdot\lambda$, where $b$ is the slit width, $D$ the slit-to-screen distance, $d$ the distance between the $m$-th maximum and the central maximum, $\lambda$ the wavelength. $D$ and $d$ are measured with least counts of $1$ cm and $1$ mm, respectively. $\lambda = 600$ nm and $m = 3$ are known precisely. With $m = 3$, $d = 5$ mm, and $D = 1$ m, find the absolute error in $b$ (in micrometers).
Reveal answer + step-by-step solution
Correct answer:76
Solution
From $b = m\lambda D/d$, the relative error is $\Delta b/b = \Delta D/D + \Delta d/d$ (since $m$ and $\lambda$ are exact). Compute $b$: $b = (3)(600 \times 10^{-9}\text{ m})(1\text{ m})/(5 \times 10^{-3}\text{ m}) = 1800 \times 10^{-9}/5 \times 10^{-3} = 360 \times 10^{-6}$ m $= 360$ μm. $\Delta D = 1$ cm $= 0.01$ m; $\Delta D/D = 0.01$. $\Delta d = 1$ mm $= 1 \times 10^{-3}$ m; $\Delta d/d = 1/5 = 0.2$. $\Delta b/b = 0.01 + 0.2 = 0.21$. $\Delta b = 0.21 \times 360 = 75.6$ μm $\approx 76$ μm. Within official acceptance band $[75, 79]$. Answer: 76 μm.
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